Given a scalene triangle $\triangle ABC$ and $H$ the orthocenter of the triangle. $P$ is a point on the Euler circle of the triangle $\triangle ABC$. The segments $BH, CH$ intesect the opposite sides $AC,AB$ at the points $E,Z$ respectively. The circumscribed circles of the triangles $\triangle EHP$ and $\triangle ZHP$ intersect the lines $CH, BH$ for a second time at the points $K, R$ respectively. Prove that when the point $P$ moves on the Euler circle of the triangle $\triangle ABC$, the line $KP$ goes through a constant point.
On this question I am very much lost. I immediately thought that $EPHK$ is inscrived and hence has all the properties of an inscribed quadrilateral. Moreover, I thought of somehow manipulating the Euler circle and somehow using the fact that it passes through the midpoints of $AB,BC,AC$ to solve the question, but none of this bared crops. Could you please explain to me how to solve this question?
Best Answer
Let $EH$ and $PK$ meet the nine point circle at points $M$ and $D$ respectively (These are different from points $E$ and $P$ respectively. ). $M$ will be the midpoint of $BH$.
Now, observe that, $MD\parallel HK$ because $\angle MDP=180-\angle MEP=180-\angle HEP=\angle HKP$. Also, notice that, $D$ is the midpoint of $BC$ because $BM=MH$ and $MD\parallel HC$. Hence, $PK$ always passes through the midpoint of $BC$, a constant point.